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Answer : The final molarity of iodide anion in the solution is 0.0508 M.
Explanation :
First we have to calculate the moles of and
.
Molar mass of KI = 166 g/mole
and,
Now we have to calculate the limiting and excess reagent.
The given chemical reaction is:
From the balanced reaction we conclude that
As, 1 mole of KI react with 1 mole of
So, 0.0178 mole of KI react with 0.0178 mole of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 1 mole of react to give 1 mole of
So, 0.0178 moles of react to give 0.0178 moles of
Thus,
Moles of AgI = Moles of anion = Moles of
cation = 0.0178 moles
Now we have to calculate the molarity of iodide anion in the solution.
Therefore, the final molarity of iodide anion in the solution is 0.0508 M.